Connecting LEDs to a 9V battery

Question

1) I have 6 4.5V 20mA LEDs and have connected them to a common 9V battery. Doing this I made 3 parallel circuits with 2 LEDs in each.

Now, I see that most pages that describe LED circuitry say that you should always have a resistor in the circuit. In my case, do I need one? Why?

This is a hobby project, but the LEDs are sealed in glue so fixing a blown bulb would be a big hassle. I'd rather have them live for as long as possible.

2) If I happened to have the same circuit as above, but with 3.4V 20mA LEDs instead, is it likely that the LEDs would blow quickly? The LEDs are of a cheap type:

http://www.ebay.com/itm/20pcs-5mm-Clear-White-LEDs-Ultra-Bright-LED-DIY-/190642214244?pt=LH_DefaultDomain_0&hash=item2c63295564#ht_3051wt_1156

Answer 1

Ideally the LEDs should be driven by a constant-current circuit. This will maintain a constant brightness and color as the battery drains, or as the LEDs heat up or cool down.

But the real world isn't ideal, so you can often get away by just using resistors. Yes, you should definitely use them. The resistors are there to limit the current through the LED and keep them from overheating and burning up. A 9V battery has a fairly high internal resistance, so you may be able to get away with two in series and no resistor, but it will be unreliable (changing to a different brand of battery could be enough to blow out the LEDs, etc.)

For the worst case of two white LEDs in series running at 20mA, the lowest forward voltage shown in your link is 3.2V, so you would have (9 - 6.4)/.02 = 130 ohms. The current is low, so a 1/4 watt resistor will be fine. Select the closest value to this you can find. Running at 20mA the LEDs will be pretty bright and this is a benefit: as the battery drains or the LED forward voltage changes, the apparent brightness probably won't change that much. Human vision is more sensitive to dim lights and it's harder to tell that a bright light has changed 10% than a dim light has changed 10%.

Answer 2

Yes, you need a series resistor. The reason is that the LED voltage isn't exactly 4.5V, and it also varies with current. Having a series resistor is the only way to be sure you supply the right amount of current. You need some voltage drop across the resistor, however, so putting two LEDs in series is out. You need 4.5V across the resistor for 20mA so

\$R = \dfrac{9V - 4.5V}{20mA} = 220\Omega\$

A 9V battery can't supply very much current, but without series resistor the current may still be too much for the LEDs. I'm currently working with white LEDs with a typical current of 20mA, which have only 25mA as maximum.

edit (re your comment)
In theory you could do without resistor, but the LEDs are never exactly 4.5V and your battery is never exactly 9V. If the battery's voltage decreases to 8.9V your LEDs will light (much) dimmer, or (much) brighter if the LEDs' voltage is 4.4V, since the current will be much higher, and maybe become too high.